A343400 Number of ways to write n as 3^x + [y^2/3] + [z^2/4], where [.] is the floor function, x is a nonnegative integer, and y and z are positive integers.

1, 2, 3, 4, 4, 5, 4, 6, 4, 9, 5, 7, 8, 6, 9, 6, 7, 9, 7, 6, 9, 7, 8, 7, 7, 10, 6, 9, 11, 9, 12, 8, 9, 14, 5, 13, 11, 8, 11, 11, 7, 13, 9, 12, 11, 9, 9, 11, 8, 12, 11, 11, 11, 6, 16, 4, 11, 12, 11, 13, 12, 6, 10, 9, 8, 17, 8, 12, 11, 10, 8, 10, 12, 10, 8, 11, 12, 12, 13, 7

Offset

1,2

Comments

Conjecture: a(n) > 0 for all n > 0.

We have verified a(n) > 0 for all n = 1..2*10^6.

The first indices n for which a(n) = 0 are 4051736, 7479656, 8592680, 9712160, 14039792, 16726256, 24914510. - Giovanni Resta, Apr 14 2021

Links

Table of n, a(n) for n=1..80.

Zhi-Wei Sun, Natural numbers represented by [x^2/a] + [y^2/b] + [z^2/c], arXiv:1504.01608 [math.NT], 2015.

Example

a(2) = 2 with 2 = 3^0 + [1^2/3] + [2^2/4] = 3^0 + [2^2/3] + [1^2/4].
a(2942) = 2 with 2942 = 3^1 + [93^2/3] + [15^2/4] = 3^7 + [44^2/3] + [21^2/4].
a(627662) = 5 with 627662 - 3^0 = [330^2/3] + [1538^2/4] = [1042^2/3] + [1031^2/4] = [1318^2/3] + [441^2/4] = [1328^2/3] + [399^2/4] = [1352^2/3] + [271^2/4].
a(1103096) = 3 with 1103096 = 3^1 + [260^2/3] + [2079^2/4] = 3^1 + [508^2/3] + [2017^2/4] = 3^9 + [328^2/3] + [2047^2/4].
a(1694294) = 3 with 1694294 = 3^8 + [860^2/3] + [2401^2/4] = 3^8 + [928^2/3] + [2367^2/4] = 3^13 + [112^2/3] + [619^2/4].

Mathematica

PowQ[n_]:=PowQ[n]=IntegerQ[Log[3, n]];
tab={}; Do[r=0; Do[If[PowQ[n-Floor[x^2/3]-Floor[y^2/4]], r=r+1], {x, 1, Sqrt[3n-1]}, {y, 1, Sqrt[4(n-Floor[x^2/3]-1)+1]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]

Crossrefs

Cf. A000290, A000244, A343387, A343391, A343397.

Sequence in context: A327715 A306574 A088527 * A372599 A030602 A133947

Adjacent sequences: A343397 A343398 A343399 * A343401 A343402 A343403

Keyword

nonn

Author

Zhi-Wei Sun, Apr 13 2021

Status

approved